User yaakov baruch - MathOverflow

How far will a random walk on the integers go?

2011-07-31T08:17:33Z

Let $S_n$ be the distance walked at time $n$ in a simple symmetric random walk on $\mathbb{Z}$ (with steps $=\pm1$).

It is well known that $\lim_{n\to\infty}(E(|S_n|)/\sqrt{n})=\sqrt{2 /\pi}$.

Can we describe the set $F$ of monotonic "growth" functions with probability 1 of being crossed by $|S_n|$ for $n>1$?

Clearly $f(x)=\sqrt{x}$ is in $F$ and I think I could prove that $g(x)=2f(x/2)$ is in $F$ whenever $f$ is. I have no idea beyond that.

Answer by Yaakov Baruch for infinite configuration of lines

2011-06-01T04:39:12Z

I can sketch a proof based on assuming this "finite" result:

A). For any pentagonal star one of the 5 triangles will have area strictly smaller than that of the central pentagon. (I think a brute force attack should yield a proof here.)

The proof of the original problem would then go as follows.

b). A) generalizes to n-agons by considering the pentagon spanned by any 5 vertices.

c). b) implies that a tiling with polygons of EQUAL AREA is not possible unless all polygons are either triangles or quadrilaterals.

d). Take one 4-tile and continue tiling next to it inside the cone enclosed by the converging lines of 2 opposite edges; we have a sequence of quadrilaterals which must end with a triangle were the the 2 lines meet. This shows that the tiling must contain a 3-tile somewhere.

e). By d) take a 3-tile and continue tiling outwards, inside each of the 3 beams generated by the lines of each pair of edges; the original 3-tile will be the first tile in each beam, but every other tile after it must be a 4-tile (build them one at a time and keep using c)). We can ignore what happens in the 3 leftover cones radiating from the 3 vertices.

f). In one of the 3 beams (which now look like ladders) take any one of the new rungs from step e) and extend it - that line will then collide with one of the other 2 beams (but cannot overlap with any of its rungs). That will cut one of the 4-tiles, creating a 5-tile.

Apologies for bumping up the question repeatedly while trying to edit my answer.

Answer by Yaakov Baruch for Blocking visibility with cylinders

2011-05-10T12:54:52Z

Here is one construction. On the horizontal xy plane place a forest of vertical cylinders of radius r<1/2 (or =1/2 if we allow contacts) centered at each point in $(\mathbb{Z} \backslash{\lbrace0\rbrace})\times(\mathbb{Z}\backslash{\lbrace0\rbrace})$; moreover place 2 similar cylinders parallel to x centered at (y=+/-1, z=0), 2 more parallel to y centered at (x=0, z=+/-1) and the last 2 parallel to z and centered at (x=+/-1, y=0). Then (0,0,0) is blocked by the forest in all directions except those in the xz and yz planes, which are blocked by the other 6 cylinders. The forest clearly does not need to be infinite and it should be easy to find an upper limit on its size.

${\bf UPDATE}$ As pointed out by Mark in a comment, the forest should be based on $(\mathbb{Z} \backslash{\lbrace0\rbrace})\times(\mathbb{Z}\backslash{\lbrace-1,0,1\rbrace})$.

Answer by Yaakov Baruch for Ideas on how to prevent a department from being shut down.

2011-04-27T23:03:38Z

I post this as an answer rather than a comment, so people can downvote: how do the originators of the petition propose the university, or the department, should pay to keep the Geometry Section? Through savings in other sections? In other less useful departments? Through increased government support? Other means? It seems to me one cannot fairly decide A is worth funding at the expense of B, without any knowledge of who/what B is.

Answer by Yaakov Baruch for Which mathematical ideas have done most to change history?

2011-03-16T09:19:17Z

$i$. (But my favorite by far is Cartesian coordinates, already mentioned by Scott Carter below.)

Answer by Yaakov Baruch for Examples of common false beliefs in mathematics.

2011-03-06T22:35:26Z

Occasionally seen on this site: if a polynomial $P:\mathbb{Q}\rightarrow\mathbb{Q}$ is injective, so must be its extension to $\mathbb{R}$.

Answer by Yaakov Baruch for Ways to prove the fundamental theorem of algebra

2010-09-20T09:06:49Z

At the risk of being highly downvoted, I can't resist reposting my comment to Andrew L's answer (or rather, question) below:

is there a purely algebraic proof that for any non constant $P$ in $\mathbb{Q}[i][X]$ and $\epsilon>0$ in $\mathbb{Q}$, there is $q$ in $\mathbb{Q}[i]$ s.t. $|P(q)|<\epsilon$?

I think the statement above is purely algebraic, but I have to admit I'm a bit uncertain as to where the boundary between algebra and analysis falls.

Answer by Yaakov Baruch for What is your favorite "strange" function?

2010-04-23T08:38:40Z

Since Mariano took my favorite already, I'll go with the stopping time function for the 3x+1 problem: http://www.ieeta.pt/~tos/3x+1.html

Answer by Yaakov Baruch for Is there a good reason why a^{2b} + b^{2a} <= 1 when a+b=1?

2010-03-15T22:57:28Z

UPDATE: this "proof" is WRONG!
We want to prove that a^(2-2a)+(1-a)^(2a)<=1 for 0<=a<=1, or for 0<=a<=1/2 because of symmetry under a -> 1-a. Set f(a)=a^(2-2a). Then we want to prove that f(a)<=1-f(1-a), but since trivially f(a)<=a in [0,1/2], we have f(a)<=a<=1-a<=1-f(1-a). QED

1 rectangle <= 4 squares

2010-02-02T12:14:16Z

Almost 25 years ago a professor at Indiana U showed me the following problem:

given a map $\mathbb{Z}^2\rightarrow\mathbb{R}$ such that the sum inside every square (parallel to the axes) is $\leq1$ in absolute value, prove that the sum inside every rectangle (parallel to axes) is $\leq4$ in absolute value.

It's fun and not too hard to prove. I believe that at the time I was able to show that the upper limit can be improved to 3.975 - but that was a lot harder and I can't say now that this is for sure the case. Also, with a computer search (old TRS 80) I produced an example containing a rectangle of area $3\frac{1}{3}$.

These are some of the questions that come to mind:

can the upper limit of 4 (or 3.975?) be improved?
can the lower limit of $3\frac{1}{3}$ be improved?
any proof/conjecture about the optimal limit?
do the results extend to maps $\mathbb{R}^2\rightarrow\mathbb{R}$, provided they are "nice" enough?
are any other generalizations of this problem possible (eg. different tilings of the plane or of other manifolds, or higher dimensions)?

Update 1 (updated 7th March 2010). See answers and comments below for examples achieving ratios as high as 181/48 = 3.7708333...!

Update 2. Here is a sketch of the proof that 4 is an upper limit. A limit of 254/67=3.79104477... is now known (see answers below), but the proof for that needs to be seeded with at least some known limit.

Given a rectangle R of size AxB, with A < B, call it "thin" if $B\geq2A$ or "fat" if $B\leq2A$ (the case B=2A is irrelevant as it is the union of 2 squares). One can draw the 4 squares on the sides of R, either facing outwards (size of envelope = (2B+A)x(2A+B)), or inwards (some spilling out on the opposite sides, size of envelope = (2B-A)xB) - call these the "big-envelope" and the "small-envelope".

Assume that R has sum 4+$\epsilon$ and that every square has sum between -1 and 1. We have 3 cases, all easy exercises to work out:

(1) for any R, the fat (2A+B)x(2B+A) big-envelope will have sum $\leq-4-3\epsilon$.

(2) for a fat R, a (2A-B)x(2B-A) sub-rectangle of the small-envelope will have sum $\leq-4-3\epsilon$;

(3) for a thin R, a thin (B-2A)x(2B-A) sub-rectangle of the small-envelope, will have sum $\geq4+3\epsilon$;

Applying any of (1)+(2), (2)+(1) or (3)+(3) produces a 3Ax3B rectangle with sum $\geq4+9\epsilon$. Iterating n times produces a $3^{n}A \times 3^{n}B$ rectangle with sum $4+9^{n}\epsilon$. Such rectangle is made of no more than AxB squares (each of size $3^{n} \times 3^{n}$) and therefore, for large enough n, one of the squares will have sum >1. $\square$

Reformulation . Given an abelian group G and a map

f: GxGxGxG -> $\mathbb{R}$ such that

1) -1<=f(a,b,c,d)<=1 if d*a=c*b (boundedness of squares),

2) f(a,b,c,d)+f(c,b,e,d)=f(a,b,e,d) for all a, b, c, d, e in G (horizontal additivity of rectangles),

3) f(a,b,c,d)+f(a,d,c,e)=f(a,b,c,e) for all a, b, c, d, e in G (vertical additivity of rectangles),

can we find a universal best bound b(G) such that -b(G) <= f <= b(G)?

All the previous work on this question amounts to the result: 181/48 <= b($\mathbb{Z}$) <= b($\mathbb{Z}x\mathbb{Z}$) <= 254/67

For non-abelian groups one could perhaps generalize the notion of "square" by lifting it from G/[G,G].

Answer by Yaakov Baruch for 1 rectangle <= 4 squares

2010-03-09T01:20:32Z

There is a new upper bound of 254/67 (= 3.79104477...).

Define 6 sets of cardinality 4:

X1={-B+A, 0, A, B}
Y1={0, A, B-A, B}

X2={-B, -B+3A, B-2A, B+A}
Y2={-2B+A, -A, B+A, 3B-A}

X3={-6B+2A, -2B-2A, 2B+3A, 6B-A}
Y3={-2B-2A, -B+6A, 2B-6A, 3B+2A}

then we already know that in the in the grid X1 x Y1 if the sum in the central AxB is $4+\epsilon$ the sum in the surrounding (2B-A)x(B-2A) is $\geq4+3\epsilon$,

similarly in the in the grid (X1 $\cup$ X2) x (Y1 $\cup$ Y2) if the sum in the central AxB is $19/5+\epsilon$ then the sum in the surrounding (2B-5A)x(5B-2A) is $\leq-19/5-21\epsilon$,

last, in the in the grid (X1 $\cup$ X2 $\cup$ X3) x (Y1 $\cup$ Y2 $\cup$ Y3) if the sum in the central AxB is $254/67+\epsilon$ then the sum in the surrounding (12B-3A)x(3B-12A) is $\geq254/67+135\epsilon$.

All of the above claims are easily verifiable with the tools already described in the previous answers and comments. I wonder if one can find sets X4 and Y4 (with 4 elements each?) to further improve the bound and maybe spot a general pattern.

Answer by Yaakov Baruch for 1 rectangle <= 4 squares

2010-03-02T20:18:10Z

Here is a summary for the $\mathbb{R}^2$ situation.

Upper limit: no improvement over the 3.8 known on $\mathbb{Z}^2$.

To create specific examples I modified the $\mathbb{Z}^2$ ones by uniformely spreading each value on the lattice over a 1x1 square:

3x1: ratio = 3-3/5 (vs. 3 on integers, 7x5 grid)
4x1: ratio = 3-1/5 (vs. 3.5 on integers, 13x16 grid)
5x1: ratio = 3-1/7 (vs. 25/7 on integers, 25x29 grid)
6x1: ratio = 3-1/23 (vs. 85/23 on integers, 31x36 grid)
8x1: ratio = 3-1/35 (vs. 26/7 on integers, 39x46 grid)
7x3: ratio = 3-1/75 (vs. 56/15 on integers, 59x57 grid)
11x1: ratio = 3-1/135 (vs. 101/27 on integers, 137x63 grid)

And the surprises are
1) that in all cases the highest sum on a square is 1.25 that of the integral case (which is the worst case scenario, but I see no obvious reason it should always be that way), and
2) we seem to be approaching 3, suggesting that for the integers perhaps the upper limit could be 3.75 and not 3.8 - but this is of course very highly speculative...

Answer by Yaakov Baruch for 1 rectangle <= 4 squares

2010-02-26T02:36:09Z

The upper bound is <3.95.

I hope the code below will show correctly...

It proves that assuming a sum >=3.95 in the central AxB rectangle of the grid ({-B,-B+A,-2A,-A,0,A,2A,B-A,B}+{0,A}) x ({-2B,-B-A,-B,-B+A,-2A,-A,0,A,2A,B-A,B,B+A,2B}+{0,B}) leads to a contradiction in a finite number of steps. 3.95 is NOT best possible for this grid, but 3.94 does not lead to a contradiction. It will be easy to refine the number, but more worthwhile is probably to search a larger grid (which starts to get slow in awk.)

awk 'BEGIN {

 A=1;
 # pick B large enough to ensure that there
 # are no accidental squares in the grid below
 B=1000;

 # setting up the grid
 x[0]=-B;       x[1]=-B+A;
 x[1]=-B+A;     x[2]=-B+2*A;
 x[3]=-2*A;     x[4]=-A;
 x[4]=-A;       x[5]=0;
 x[5]=0;        x[6]=A;
 x[6]=A;        x[7]=2*A;
 x[7]=2*A;      x[8]=3*A;
 x[9]=B-A;     x[10]=B;
 x[10]=B;       x[11]=B+A;
 M=11;

 y[0]=-2*B;     y[2]=-B;
 y[1]=-B-A;     y[5]=-A;
 y[2]=-B;       y[6]=0;
 y[3]=-B+A;     y[7]=A;
 y[4]=-2*A;     y[9]=B-2*A;
 y[5]=-A;       y[10]=B-A;
 y[6]=0;        y[11]=B;
 y[7]=A;        y[12]=B+A;
 y[8]=2*A;      y[13]=B+2*A;
 y[10]=B-A;     y[14]=B+B-A;
 y[11]=B;       y[15]=B+B;
 y[12]=B+A;     y[16]=B+B+A;
 y[15]=2*B;     y[17]=3*B;
 N=17;

 for(i=0; i<=M; i++)
     for(j=i; j<=M; j++)
         for(k=0; k<=N; k++)
             for(l=k; l<=N; l++)
                 # 0 sum for degenerate rectangles
                 if(i==j || k==l) {
                     lo[i,j,k,l]=0;
                     hi[i,j,k,l]=0;
                 }                   
                 # squares
                 else if(x[j]-x[i]==y[l]-y[k]) {
                     lo[i,j,k,l]=-1;
                     hi[i,j,k,l]=1;
                 }
                 # other rectangles
                 else {
                     lo[i,j,k,l]=-4;
                     hi[i,j,k,l]=4;
                 }

 # central rectangle: assume its sum is >=3.95
 lo[5,6,6,11]=3.95;

 iter=10000;
 active=1;
 while(iter-- && active) {
     active=0;

     # traverse all possible combinations of 1 rectangle split into 4
     for(i=0; i<M; i++)
         for(j=i+1; j<=M; j++)
             for(k=0; k<N; k++)
                 for(l=k+1; l<=N; l++)
                     for(m=i; m<j; m++)
                         for(n=k; n<l; n++) {
                             lo0=lo[i,j,k,l];
                             lo1=lo[i,m,k,n];
                             lo2=lo[m,j,k,n];
                             lo3=lo[i,m,n,l];
                             lo4=lo[m,j,n,l];
                             hi0=hi[i,j,k,l];
                             hi1=hi[i,m,k,n];
                             hi2=hi[m,j,k,n];
                             hi3=hi[i,m,n,l];
                             hi4=hi[m,j,n,l];

                             # 3rd argument in max() and min() funtions
                             # is for printing purposes only...
                             lo0=max(lo0, lo1+lo2+lo3+lo4, 0);
                             hi0=min(hi0, hi1+hi2+hi3+hi4, 0);
                             lo1=max(lo1, lo0-hi2-hi3-hi4, 1);
                             lo2=max(lo2, lo0-hi1-hi3-hi4, 2);
                             lo3=max(lo3, lo0-hi1-hi2-hi4, 3);
                             lo4=max(lo4, lo0-hi1-hi2-hi3, 4);
                             hi1=min(hi1, hi0-lo2-lo3-lo4, 1);
                             hi2=min(hi2, hi0-lo1-lo3-lo4, 2);
                             hi3=min(hi3, hi0-lo1-lo2-lo4, 3);
                             hi4=min(hi4, hi0-lo1-lo2-lo3, 4);

                             if(lo0>hi0 || lo1>hi1 || lo2>hi2 || lo3>hi3 || lo4>hi4) {
                                 print "CONTRADICTION AT", i,m,j,k,n,l;
                                 exit;
                             }

                             lo[i,j,k,l]=lo0;
                             lo[i,m,k,n]=lo1;
                             lo[m,j,k,n]=lo2;
                             lo[i,m,n,l]=lo3;
                             lo[m,j,n,l]=lo4;
                             hi[i,j,k,l]=hi0;
                             hi[i,m,k,n]=hi1;
                             hi[m,j,k,n]=hi2;
                             hi[i,m,n,l]=hi3;
                             hi[m,j,n,l]=hi4;
                         }
 }
 print "FINISHED OK";
}

function max(s,t, where) {

if(s<t) {
    print "lo=" t, "for", i,m,j,k,n,l, "(" where ")";
    active=1;
    s=t;
}
return(s);
}

function min(s,t, where) {

if(s>t) {
    print "hi=" t, "for", i,m,j,k,n,l, "(" where ")";
    active=1;
    s=t;
}
return(s);
}
'

Answer by Yaakov Baruch for Is a function which is finitely multiple-valued in each variable separately, also finitely multiple-valued in all its variables jointly?

2010-02-11T17:07:56Z

Engineer a function that has n>=2 values exactly in the set $S_n \backslash S_{n+1}$ where $S_n$={(x,y) | x>n, y>n, (x-n)*(y-n)>1}, and 1 value elsewehere. That should work as a counterexample, as far as $C^{\infty}$ functions go - analytic, not sure

Answer by Yaakov Baruch for What's your favorite equation, formula, identity or inequality?

2010-02-08T14:14:11Z

Trivial as this is, it has amazed me for decades:

$(1+2+3+...+n)^2=(1^3+2^3+3^3+...+n^3)$

Answer by Yaakov Baruch for Is it possible to capture a sphere in a knot?

2009-12-28T11:42:14Z

Since Anton's beautiful solution makes use of the symmetry of the sphere, I wonder how similar results could be proven, or counterexamples given, for any other convex shape, including 2-dimensional ones - i.e. infinitely thin 3-dimensional objects. I can't figure a way to tie a cube or a square, but it seems that an equilateral triangle could be tied by 3 connected loops starting from some knot at the center and going over each vertex (forming 3 equilateral triangles 1/3 the size of the original).

(I would have liked to just leave this post as a comment to Anton's proof, but I'm not allowed to do that. Should it perhaps be a new question?)

Answer by Yaakov Baruch for Is it possible to capture a sphere in a knot?

2009-12-08T18:28:57Z

Reid, excellent proof. It works for the cube too and even kills the icosahedron. In this last case I don't know what the angles of a triangle are exactly, so let's just say 60+ each. Then joining each vertex to the center gives smaller triangles with 120, 30+ and 30+ degree angles and it's then clear that shrinking a triangle towards its center will again reduce the total length. More generally, your idea seems to kill the possibility any small enough face with 3, 4 or 5 edges, and one of those is certainly needed somewhere! My bet is now confidently on the sphere always escaping.

Answer by Yaakov Baruch for Is it possible to capture a sphere in a knot?

2009-12-08T11:38:23Z

It seems that both the 2-agon and the octahedron (which after all is a collection of 3 somewhat constrained 2-agons) can be shrunk off the sphere, but with 0 derivative at the start, which means that they come close. Perhaps the icosahedron could work (as for tying 5 edges into one vertex, a tangle of simple knots should do). But notice that inflating a triangle (even though it's fairly flat on an icosahedron) will trade off with shrinking not 3 (tetrahedron) or 6 (octahedron) other edges, but 9 of them. Such trade off would increase the total length on a plane, but on the sphere it depends on the curvature, i.e. the size of the triangles (similarly to what I pointed out re. the dodecahedron in an earlier post).

I suspect the just like no hexagon at all can be allowed on a graph tessellating the sphere, so no square or pentagon can be allowed to belong to any vertex other than a 3-edged one. If that's the case, probably the cube (which no one seems to have ruled out yet), icosahedron and dodecahedron are the only candidates for graphs with strictly locally minimal total length.

Other tessellations of the sphere with mixed polygons probably shouldn't work either. For example, considering a spherical triangular prism (2 squares and a triangle meeting at each of the 3 vertices): while inflating a triangle is harder than on a spherical tetrahedron, inflating a square will be easier than on a spherical cube (whose square is less curved). Similarly for a pentagonal prism: advantageous from the point of view of its 5 squares, but not from that of its 2 pentagons (when compared with the dodecahedron).

Very nice problem!

Answer by Yaakov Baruch for Is it possible to capture a sphere in a knot?

2009-12-08T01:53:21Z

Adding to Zeb's proof that the tetrahedron can be deformed, one should notice that any tassellation of the sphere containing at least one hexagon (fullerene type) won't be rigid either. In fact already in the plane the regular hexagon can be inflated (at the expense of the 6 outgoing rays) with no change in total length, much more so on a positively curved surface. Like Zeb, I'm also skeptical about the cube. Perhaps the dodecahedron has a chance, since its pentagons are quite a bit flatter than the faces of either the cube or tetrahedron (but then the inflation procedure is also cheaper than for the triangle or the square).

Splitting pythagorean triples

2009-10-18T17:17:41Z

Can one partition the set of positive integers into finitely many pythagorean-triple-free subsets? If so, what is the smallest number of such subsets? Taking a wild guess, I would be least surprised if the answer were 3.

Notice that the 2 subsets of integers such that highest power of 5 that divides them is a) even b) odd manage to split most primitive triples, plus all the multiples of those.

Notice also that Schur proved the positive integers cannot be split into any finite number of sum-free subsets (i.e. no finite partition can split all power-of-1 Fermat triples), while Fermat's theorem proves that all power-of-n (n>2) triples can be split by the trivial partition into 1 set.

Edit: Since this turns out to be a known open problem, we're adding the tag [open-problem] and converting this question to community wiki. The idea is to have a separate answer for each possible approach to solving this problem. If you have some additional insight or a reference to contribute to an answer, you only need 100 rep to do so. We're still figuring out exactly how to handle open problems on MO. The discussion is happening on this meta.mathoverflow.net thread.

Answer by Yaakov Baruch for Splitting pythagorean triples

2009-10-19T21:22:35Z

thank you Kevin and everyone else. I've taking stabs at this pblm for years... at least I don't feel so bad I couldn't get anywhere. YB

Comment by Yaakov Baruch

2012-03-14T18:53:44Z

I think the comment stream to the answer below is relevant to this discussion: mathoverflow.net/questions/28806/… especially the final comments by Joel David Hamkins and Carl Mummert

Comment by Yaakov Baruch

2012-03-13T13:07:48Z

not to be mean or anything... Diab is NOT a toddler... and is this really a suitable question for MO??

Comment by Yaakov Baruch

2011-10-31T18:24:55Z

@Peter: as far as an injective $p$ in $n$ variables, it seems to be almost certainly possible - see the 3rd comment from top.

Comment by Yaakov Baruch

2011-08-02T04:28:10Z

Why don't you write $\frac{2\pi^4}{315} \prod_n^\infty{(1-\frac{1}{p_n-p_n^2})}$?

Comment by Yaakov Baruch

2011-08-01T17:23:49Z

I agree with the vote to close, but I would like to first select Leonid Petrov's answer if he posted it as such.

Comment by Yaakov Baruch

2011-07-31T11:37:45Z

@Leonid: I think your comment is the answer to the question (or at least the INTENDED question).

Comment by Yaakov Baruch

2011-07-31T11:23:10Z

Gerry Myerson: yes, I edited accordingly. @Ricky Demer: I replaced "hit" with "crossed" - I hope that clarifies that I'm not looking for equality of two functions, but for $f$'s such that $\limsup |S_n|/f(n) \ge 1$ with probability 1

Comment by Yaakov Baruch

2011-06-18T23:33:35Z

Re. the pentagonal star, one can wiggle around any one of the 5 lines in such a way that the pentagon and 3 triangles keep the same area, while the other 2 triangles are equalized. Therefore one can assume all 5 triangles to have the same area. Forcing this condition on the pentagon with vertices (1,0), (0,0), (0,1), (a,b), (c,d) can easily be seen to determine a, b, c and d. Therefore, up to affine transformation, the unique star whose triangles have equal area is the regular star.

Comment by Yaakov Baruch

2011-06-10T12:13:55Z

Also see mathoverflow.net/questions/67117/… and in particular Vladimir Dotsenko's answer, where other similar relations are listed, including $2P_3^2=P_7+P_5$.

Comment by Yaakov Baruch

2011-06-05T08:50:56Z

@John: copy the comment, delete it, paste, edit and re-post it - but quickly enough that someone's reply won't sneak in!

Comment by Yaakov Baruch

2011-05-31T19:45:40Z

A) Is it always true that for a pentagonal star, one of the 5 triangles has area strictly smaller than the central pentagon? (Can assume the pointy tips lie on a circle if it helps, or else that 2 sides are orthogonal and of length 1.) B) that should prove that any equal area tiling could can only consist of triangles and quadrilaterals... C) but then what? D) Love the question! Seemed like such an easy, overconstrained no-brainer!

Comment by Yaakov Baruch

2011-05-26T18:54:35Z

is Moebius^2 homeomorphic to cylinder^2?

Comment by Yaakov Baruch

2011-05-12T18:44:27Z

Since it's now closed, I de-down-voted.

Comment by Yaakov Baruch

2011-05-12T18:26:20Z

I reluctantly voted down because I just don't think it's right for such a question to almost monopolize the top of the list for 2 days.

Comment by Yaakov Baruch

2011-05-11T07:51:25Z

Awesome graphics! Yaakov "There Is No Light At The End Of The Tunnel" Baruch.